Let $f(x)=\frac{3 \sin (x)}{1+\cos (x)}$. Find The Following:1. $f^{\prime}(x)=$ \$\square$[/tex\]2. $f^{\prime}(3)=$ $\square$

Feb 26, 2025 by ADMIN 137 views

Finding Derivatives and Evaluating Functions: A Mathematical Exploration

In this article, we will delve into the world of calculus and explore the concept of derivatives and function evaluation. We will examine a given function, $f(x)=\frac{3 \sin (x)}{1+\cos (x)}$ , and find its derivative, $f^{\prime}(x)$ . Additionally, we will evaluate the derivative at a specific point, $f^{\prime}(3)$ .

The derivative of a function represents the rate of change of the function with respect to its input variable. It is a fundamental concept in calculus and has numerous applications in various fields, including physics, engineering, and economics.

To find the derivative of a function, we can use various techniques, such as the power rule, product rule, and quotient rule. In this case, we will use the quotient rule to find the derivative of $f(x)$ .

Quotient Rule

The quotient rule states that if we have a function of the form $f(x)=\frac{g(x)}{h(x)}$ , then the derivative of $f(x)$ is given by:

f^{\prime}(x)=\frac{h(x)g^{\prime}(x)-g(x)h^{\prime}(x)}{(h(x))^2}

Finding the Derivative of $f(x)$

Using the quotient rule, we can find the derivative of $f(x)$ as follows:

f^{\prime}(x)=\frac{(1+\cos (x))\frac{d}{dx}(3 \sin (x))-(3 \sin (x))\frac{d}{dx}(1+\cos (x))}{(1+\cos (x))^2}

f^{\prime}(x)=\frac{(1+\cos (x))(3 \cos (x))-3 \sin (x)(-\sin (x))}{(1+\cos (x))^2}

f^{\prime}(x)=\frac{3 \cos (x)+3 \cos^2 (x)+3 \sin^2 (x)}{(1+\cos (x))^2}

f^{\prime}(x)=\frac{3 \cos (x)+3}{(1+\cos (x))^2}

f^{\prime}(x)=\frac{3(1+\cos (x))}{(1+\cos (x))^2}

f^{\prime}(x)=\frac{3}{1+\cos (x)}

Evaluating the Derivative at a Specific Point

Now that we have found the derivative of $f(x)$ , we can evaluate it at a specific point, $x=3$ . To do this, we substitute $x=3$ into the derivative:

f^{\prime}(3)=\frac{3}{1+\cos (3)}

Using a calculator or a trigonometric table, we can find the value of $\cos (3)$ :

\cos (3) \approx -0.9903

Substituting this value into the derivative, we get:

f^{\prime}(3)=\frac{3}{1-0.9903}

f^{\prime}(3)=\frac{3}{0.0097}

f^{\prime}(3) \approx 309.79

In this article, we have explored the concept of derivatives and function evaluation. We have found the derivative of a given function, $f(x)=\frac{3 \sin (x)}{1+\cos (x)}$ , using the quotient rule. We have also evaluated the derivative at a specific point, $x=3$ . The results demonstrate the importance of derivatives in understanding the behavior of functions and their applications in various fields.

This article has provided a basic introduction to the concept of derivatives and function evaluation. However, there are many more topics to explore in this field, including:

Higher-order derivatives: The derivative of a function can be taken multiple times to obtain higher-order derivatives.
Multivariable calculus: Derivatives can be extended to functions of multiple variables, which is essential in many fields, including physics and engineering.
Applications of derivatives: Derivatives have numerous applications in various fields, including physics, engineering, economics, and computer science.

These topics will be explored in future articles, providing a deeper understanding of the concept of derivatives and their applications.

Calculus: Michael Spivak, W.W. Norton & Company, 2008.
Trigonometry: Charles P. McKeague, Brooks/Cole Publishing Company, 2005.
Derivatives: James Stewart, Brooks/Cole Publishing Company, 2008.

Note: The references provided are for general information and are not specific to the topic of this article.
Derivatives and Function Evaluation: A Q&A Guide

In our previous article, we explored the concept of derivatives and function evaluation. We found the derivative of a given function, $f(x)=\frac{3 \sin (x)}{1+\cos (x)}$ , using the quotient rule and evaluated it at a specific point, $x=3$ . In this article, we will answer some frequently asked questions (FAQs) related to derivatives and function evaluation.

Q: What is a derivative?

A: A derivative represents the rate of change of a function with respect to its input variable. It is a fundamental concept in calculus and has numerous applications in various fields, including physics, engineering, and economics.

Q: How do I find the derivative of a function?

A: There are several techniques to find the derivative of a function, including the power rule, product rule, and quotient rule. The quotient rule is used to find the derivative of a function of the form $f(x)=\frac{g(x)}{h(x)}$ .

Q: What is the quotient rule?

A: The quotient rule states that if we have a function of the form $f(x)=\frac{g(x)}{h(x)}$ , then the derivative of $f(x)$ is given by:

f^{\prime}(x)=\frac{h(x)g^{\prime}(x)-g(x)h^{\prime}(x)}{(h(x))^2}

Q: How do I evaluate the derivative at a specific point?

A: To evaluate the derivative at a specific point, we substitute the value of the input variable into the derivative. For example, to evaluate the derivative of $f(x)=\frac{3 \sin (x)}{1+\cos (x)}$ at $x=3$ , we substitute $x=3$ into the derivative:

f^{\prime}(3)=\frac{3}{1+\cos (3)}

Q: What are some common applications of derivatives?

A: Derivatives have numerous applications in various fields, including:

Physics: Derivatives are used to describe the motion of objects, including velocity and acceleration.
Engineering: Derivatives are used to design and optimize systems, including electrical and mechanical systems.
Economics: Derivatives are used to model economic systems and make predictions about future economic trends.
Computer Science: Derivatives are used in machine learning and artificial intelligence to optimize algorithms and make predictions.

Q: What are some common mistakes to avoid when working with derivatives?

A: Some common mistakes to avoid when working with derivatives include:

Not checking the domain of the function: Make sure the function is defined for the input value.
Not using the correct derivative rule: Use the correct derivative rule for the function.
Not evaluating the derivative at the correct point: Make sure to substitute the correct value into the derivative.

In this article, we have answered some frequently asked questions related to derivatives and function evaluation. We have provided a brief overview of the concept of derivatives and their applications, as well as some common mistakes to avoid when working with derivatives. We hope this article has been helpful in understanding the concept of derivatives and their applications.